# Two mercury droplets of radii 0.1 cm and 0.2 cm collapse into one single drop. What amount of energy is released? The surface tension of mercury .

Surface tension energy,

As we can infer from the law of conservation of mass, the volume of a drop V = v1 + v2

$r_1 = 0.1 cm = 0.001 m$

$r_2 = 0.2 cm = 0.002 m$

$\Delta A = 4 \pi r_1^{2} + 4 \pi r_2^{2} - 4\pi R^{2}$

$\Delta A = 4 \pi \left [r_1^{2} + r_2^{2}- R^{2} \right ]$

The new bigger drop has a radius of R,

$\frac{4}{3}\pi R^{3}=\frac{4}{3}\pi \left [ r_1^{3}+r_2^{3} \right ]$

$R^{3}= r_1^{3}+r_2^{3}$

Now,

Finally, the area will be smaller, when energy is released as the bigger drop is formed from smaller drops.

## Most Viewed Questions

### Preparation Products

##### Knockout NEET 2024

Personalized AI Tutor and Adaptive Time Table, Self Study Material, Unlimited Mock Tests and Personalized Analysis Reports, 24x7 Doubt Chat Support,.

₹ 40000/-
##### Knockout NEET 2025

Personalized AI Tutor and Adaptive Time Table, Self Study Material, Unlimited Mock Tests and Personalized Analysis Reports, 24x7 Doubt Chat Support,.

₹ 45000/-
##### NEET Foundation + Knockout NEET 2024

Personalized AI Tutor and Adaptive Time Table, Self Study Material, Unlimited Mock Tests and Personalized Analysis Reports, 24x7 Doubt Chat Support,.

₹ 54999/- ₹ 42499/-
##### NEET Foundation + Knockout NEET 2024 (Easy Installment)

Personalized AI Tutor and Adaptive Time Table, Self Study Material, Unlimited Mock Tests and Personalized Analysis Reports, 24x7 Doubt Chat Support,.

₹ 3999/-